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2

It seems the following. I found the original problem (second picture, Problem 15). Unfortunately, the problem expression is long and complex and there are misprints in your version. The right formulation should be: $$1949(xyzuvw+xyzu+xyzw+xyvw+xuvw+zuvw+xy+xu+xw+zu+zw+vw+1)= 2004(yzUvw+yzu+yzw+YVW+uvw+y+u+w).$$ The problem was solved by thepiano (here ...


1

Here's the info on this continued fraction. http://mathworld.wolfram.com/ContinuedFractionConstant.html


5

This answer contains computer-assisted algebra bashing. Pencil wielding mathematicians be warned. Let $$f_k(n)=\frac{1}{1+\frac{n^2}{1+\frac{1}{\frac{^\ddots_1}{1+n^2}}}}$$ where there are $2k$ horizontal fraction bars. Then, we have $$f_2(n)=\frac{1}{1+\frac{n^2}{1+\frac{1}{1+\frac{1}{1+n^2}}}}=\frac{\frac{1}2}{n^2+1}+\frac{\frac{3}2}{n^2+3}$$ ...


5

This is one of the formulas that Ramanujan sent Hardy in a letter. It is one of three that Hardy says "defeated me completely; I had never seen anything in the least like them before. A single look at them is enough to show that they could only be written down by a mathematician of the highest class. They must be true because, if they were not true, no ...


10

This is again only a part of a proof, because I believe I made a mistake somewhere along the line. Anyway, we'll start from this equation $$ \frac{1}{1+\frac{n^2}{1+\frac{1}{1+\frac{\ddots}{1+\frac{1}{1+\frac{1}{1+n^2}}}}}}=\frac{F_{k-1}\:n^2+F_{k}}{F_{k-2}\:n^4+2F_{k-1}\:n^2+F_{k}} $$ where k is equal to the number of horizontal lines on the left hand side. ...


4

This is not a full answer but a partial result. You can prove by induction that the described fraction of yours with $k$ horizontal lines is equal to $\frac{F_{k-1}n^2+F_{k}}{F_{k-2}n^4+2F_{k-1}n^2+F_k}$ with $k\ge3$. Perhaps with partial fractioning, you can compute the series using the well known result: $$ \sum_{n=1}^{\infty}\frac{1}{n^2+z^2}=\frac{\pi ...


1

The continued fraction representation $x=[a_0;a_1,a_2,\ldots]$ is unique when $x$ is irrational. If $x$ is rational, there are exactly two continued fraction representations: $x=[a_0;a_1,\cdots,a_n]=[a_0,a_1,\cdots,(a_{n}-1),1]$, so it's easy to see the question-mark function is indeed independent of continued fraction representation.


2

The left hand side is $ \tanh \frac{\pi}{2} $ so it may be worth looking at a continued fraction for $ \tanh x $. It looks like a special case of Gauss's continued fraction (I can't link to it directly but there is a Wikipedia page on it). Also, Ramanujan is not mentioned.


2

For $x,y > 0$, we have for all $i$ that: \begin{align*} d \left(\frac1{a_i+x}, \frac1{a_i+y}\right) &= \left|\frac{(a_i+y) - (a_i+x)}{(a_i+x)(a_i+y)}\right| \\ &= \frac{d(x,y)}{(a_i+x)(a_i+y)}\\ &\le a_i^{-2} d(x,y) \end{align*} It follows that $d(x_{n+1},x_n) \le \prod\limits_{i=1}^k a_i^{-2} d(x_n, x_{n-1})$ so that this product may serve ...


1

While the question being asked can be settled using ergodicity of the Gauss map and the ergodic theorem, as I explained in comments to the question, it can also be explained using only the ergodicity alone (which is itself a nontrivial feature of the Gauss map, so it's not like we're getting something for free). I'll state the result in a general context, as ...



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